## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 33

#### Answer

Any of: $4\sin x\cos x(1-2\sin^{2}x)$ $4\sin x\cos x(\cos^{2}x-\sin^{2}x)$ $4\sin x\cos x(2\cos^{2}x- 1)$

#### Work Step by Step

$\sin 4x=\sin[2(2x)] =$ ... Double-Angle identity $...\sin 2A=2\sin A\cos A$ $=2\sin 2x\cos 2x$ ... Double-Angle ... $\sin 2A=2\sin A\cos A$ ... $\cos 2A=1-2\sin^{2}A$ $=2(2\sin x\cos x)(1-2\sin^{2}x)$ $=4\sin x\cos x(1-2\sin^{2}x)$ (answers may differ if other double angle identities for cosine are chosen, $\cos 2A=\cos^{2}A-\sin^{2}A$, $\cos 2A=2\cos^{2}A- 1$) Answer: any of $4\sin x\cos x(1-2\sin^{2}x)$ $4\sin x\cos x(\cos^{2}x-\sin^{2}x)$ $4\sin x\cos x(2\cos^{2}x- 1$

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